The system of linear equations $x + y + z = 6, x + 2 y + 3 z = 10, x + 2 y + 2 z = 4$ will have

non trivial solution

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

trivial solution

No worries! We‘ve got your back. Try BYJU‘S free classes today!

infinite solutions

No worries! We‘ve got your back. Try BYJU‘S free classes today!

no solutions

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
non trivial solution

Once you express the given equations in the form of matrices then the determinant $D = - 1, D_{1} = - 8, D_{2} = 8, D_{3} = - 6$ . Here $D! = 0$ , none of $D_{1}, D_{2}, D_{3}$ are zeros. So according to Cramer's rule there exists non trivial solutions and the solutions being $x = \frac{D_{1}}{D}, y = \frac{D_{2}}{D}, z = \frac{D_{3}}{D}$

Suggest Corrections

Similar questions

x + y + z = 6, x + 2 y + 3 z = 10

x + 2 y + a z = b

The system of linear equations x+y+z=6, x+2y+3z=10 and x+2y+az=b has no solutions when …….

Number of solutions for the system of linear equations, will be,

x + y + z =6

x + 2y + 3z = 14

x + 4y + 2z =30

Number of solutions for the system of linear equations, $x + y + z = 6, x + 2 y + 3 z = 14,$ and $x + 4 y + 2 z = 30$ are

Join BYJU'S Learning Program